When we take two neighboring sides of the square and the diagonal of the square, we have a right triangle. If each side of the square is √6, then we can use the pythagorean theorem to find the length of the diagonal:
$$\sqrt{6}^2\ +\ \sqrt{6}^2=d^2$$ $$12=d^2$$ $$d\ =\ \sqrt{12}$$ where d is the diagonal. This means that each side of the triangle is √12.
To find the area of an equilateral triangle, we can use the area of an equilateral triangle area formula $$A\ =\ \frac{\sqrt{3}}{4}a^2$$ where a is the length of one side of the triangle (√12). However, if we don't know this formula, we can imagine that we divide the equilateral triangle down the middle, creating two 30-60-90 right triangles with a hypotenuse of √12, a base of √12/2, and a height that is the same as the height of the full equilateral triangle. Given that the side lengths of a 30-60-90 triangle are x -- x√3 -- 2x, we know that x must be √12/2, 2x must be √12, making our third side length (the height) √3*√12/2. (Note: if we didn't recognize this as a 30-60-90 triangle, we could have used the pythagorean theorem again to solve for the height.)

So then the area of the full equilateral triangle is 1/2bh, where h = √3*√12/2 and b = √12:
$$A=\frac{1}{2}\left(\sqrt{12}\right)\left(\frac{\sqrt{3}\sqrt{12}}{2}\right)$$ $$A=\frac{1}{4}\left(12\right)\left(\sqrt{3}\right)$$ $$A=3\sqrt{3}$$

swerve wrote:

If the diagonal of a square of side √6 is the side of an equilateral triangle, what is the area of this triangle?

$$A.\ \frac{3}{8}\sqrt{3}$$
$$B.\ \frac{3}{4}\sqrt{3}$$
$$C.\ \frac{3}{2}\sqrt{3}$$
$$D.\ 2\sqrt{6}$$
$$E.\ 3\sqrt{3}$$

Let's determine the diagonal of the square first. Since diagonal = side x √2, the diagonal = √6 x √2 = √12.

Since the diagonal of the square = side of the the equilateral triangle and the area of an equilateral triangle with side s is given by A = (s^2 √3)/4, we have

A = [(√12)^2 √3]/4

A = 12√3/4

A = 3√3

Answer: E